The KE of an electron emitted from tunsten surface is 3.06 eV. What voltae would be required to bring an electron to rest?

KE of electron emitted frm tungsten surface is 3.06*1.602*10^-19 J = 4.902*10^-19

Let electron has been accelerated from rest through a potential difference of  V

  KE = qV = 1.6*10^-19 *V = 4.902*10^-19

         = 3.064 Volts

---

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and
we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch